What is the integral of ((e^{-1/x}))/(x^2) ?

The integral of ((e^{-1/x}))/(x^2) is e^{-1/x}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of ((e^{-1/x}))/(x^2)

\int \frac{( e ^{- \frac{1}{x}} )}{x ^{2}} d x

e^{- \frac{1}{x}} + C

Solution steps

\int \frac{e ^{- \frac{1}{x}}}{x ^{2}} d x

Apply u-substitution

: \int e^{u} d u

= \int e^{u} d u

Use the common integral:

\int e^{u} d u = e^{u}

= e^{u}

Substitute back

u = - \frac{1}{x}

= e^{- \frac{1}{x}}

Add a constant to the solution

= e^{- \frac{1}{x}} + C

\int \frac{cos ( x )}{( sin ( x ) - 1 ) ^{2}} d x

\frac{\partial}{\partial x} (x^{2} + y^{2} (1 - x)^{3})

\frac{\partial}{\partial x} (\frac{x}{1 - x})

\int \frac{3 x - 4}{( x - 2 ) ( x ^{2} + 1 )} d x

\int \frac{x ^{2} - 16}{x - 4} d x

What is the integral of ((e^{-1/x}))/(x^2) ?
The integral of ((e^{-1/x}))/(x^2) is e^{-1/x}+C